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Op-Amp Simulation – Part II EE/CS 5720/6720

This assignment continues the simulation and characterization of a simple operational amplifier. Turn in a copy of this assignment with answers in the appropriate blanks, and Cadence printouts attached. All problems to be turned in are marked in boldface.

For the following problems, use the two-stage op amp you simulated in the previous assignment, using the same value of C C and the same lead compensation transistor you arrived at. For all simulations below, load the amplifier with R L = 1M Ω in parallel with C L = 30pF.

1. Common-mode gain; CMRR

Common-mode gain measures how much the output changes in response to a change in the common-mode input level. Ideally, the common-mode gain of an op amp is zero; the amplifier should ignore the common-mode level and amplify only the differential-mode signal. Let’s measure the common-mode gain of our op amp.

In order to measure the common-mode gain in the open-loop condition, we have to once again “balance” our high-gain op amp very carefully to keep V OUT ≈ 0, just like we did in the last assignment when we measured the transfer function. Remember, we do this by adding a dc voltage source V OS in series with one of the inputs. This voltage source is set to the input offset voltage so that if no other signal is present, the output voltage will be approximately zero. Now, with this adjustment in place, we tie the two inputs together and apply an ac signal v IN , as shown below.

L

v OUT

v IN V OS

Plot the common-mode gain (in dB) transfer function of the op amp over the frequency range 1Hz – 100MHz. Plot at least 50 points per decade of frequency for good resolution. Turn in this plot.

What is the common-mode gain at 10 Hz? ____________________

What is the common-mode gain at 100 kHz? ____________________

An important figure of merit in op amp design is the common-mode rejection ratio , or CMRR . CMRR is defined as the differential-mode gain divided by the common-mode gain. (Remember, if you express your gains in the logarithmic units of dB, subtraction is

equivalent to division.) For example, if a particular amplifier has a differential gain of 80 dB at 100 Hz and a common-mode gain of 10 dB at the same frequency, then the amplifier’s CMRR at 100 Hz is 70 dB. Ideally, an amplifier should have infinite CMRR. Practically, most designers try to get CMRR > 60 dB, though some applications may required much higher values.

Disconnect the negative input of the op amp from v IN and connect it back to ground. Measure the differential-mode gain (in dB) transfer function of the op amp over the frequency range 1Hz – 100MHz. (This is the same measurement you did in the last assignment.) Plot at least 50 points per decade of frequency for good resolution. Turn in this plot.

What is the CMRR at 10 Hz (in dB)? ____________________

What is the CMRR at 100 kHz (in dB)? ____________________

2. Alternate method for measuring open-loop transfer function

The previous method we used for measuring transfer functions can become slow and tedious if we often make changes to our op amp that affect its dc operating point, because this requires re-measuring the small dc offset voltage, which will have changed. Luckily, changing the value of C C has no affect on the dc bias point, so we haven’t had to repeat the dc offset measurements yet. However, if we make any changes to transistor sizes or bias currents, we would have to repeat the dc sweep to find V OS before measuring the transfer function again.

It turns out there is an easier way to measure open-loop transfer functions that does not require us to measure V OS and then “balance” the open-loop op amp. The measurement configuration is shown below.

L

v OUT v IN

First, we make R >> R L so that this resistor has no significant loading effect on the op amp. Let’s set R = 100M Ω in our simulation.

Here’s how this configuration works: At dc (and very low frequencies), C is basically an open circuit. Since no current flows into the op amp’s inputs (or through C ), the current through R is zero. That means the voltage drop across R is also zero, so the voltage at the negative input of the op amp is equal to the output voltage. Thus, at very low frequencies, the op amp is configured at a unity-gain buffer, so v OUT ≅ v IN . If we make the dc value of v IN = 0, then our output is where we want it to be.